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(C) When a ball is dropped from the top of a tower,the initial velocity $u = 0$.The distance covered in the $n^{th}$ second is given by $h_n = u + \fr

Vedclass · Accepted Answer

$1 : 3 : 5$

Vedclass · Answer

(C) When a ball is dropped from the top of a tower,the initial velocity $u = 0$.The distance covered in the $n^{th}$ second is given by $h_n = u + \frac{g}{2}(2n - 1)$.Since $u = 0$,$h_n \propto (2n - 1)$.For $n = 1, 2, 3$,the distances are $h_1 : h_2 : h_3 = (2(1)-1) : (2(2)-1) : (2(3)-1) = 1 : 3 : 5$.The work done by gravity is $W = mgh$.Since $m$ and $g$ are constant,the ratio of work done is equal to the ratio of distances covered:$W_1 : W_2 : W_3 = h_1 : h_2 : h_3 = 1 : 3 : 5$.

$A$ ball is dropped from the top of a tower. What is the ratio of the work done by the force of gravity during the $1^{st}$,$2^{nd}$,and $3^{rd}$ seconds?

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